Complex Numbers And Quadratic Equations question 151
Question: If the complex number $ z_1,z_2 $ the origin form an equilateral triangle then $ z_1^{2}+z_2^{2}= $ [IIT 1983]
Options:
A) $ z_1z_2 $
B) $ z_1\overline{z_2} $
C) $ \overline{z_2}z_1 $
D) $ |z_1{{|}^{2}}=|z_2{{|}^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ OA,OB $ be the sides of an equilateral $ \Delta OAB $ and let $ OA,OB $ represent the complex numbers or vectors $ z_1,z_2 $ respectively. From the equilateral $ \Delta OAB,\overrightarrow{AB}=z_2-z_1 $
$ \therefore $ $ arg( \frac{z_2-z_1}{z_2} )=arg(z_2-z_1)-argz_2=\pi /3 $ and $ arg( \frac{z_2}{z_1} )=arg(z_2)-arg(z_1)=\frac{\pi }{3} $ Also $ | \frac{z_2-z_1}{z_2} |=1=| \frac{z_2}{z_1} | $ , since triangle is equilateral. Thus the vectors $ \frac{z_2-z_1}{z_2} $ and $ \frac{z_2}{z_1} $ have same modulus and same argument, which implies that the vectors are equal, that is $ \frac{z_2-z_1}{z_2}=\frac{z_2}{z_1} $
Þ $ z_1z_2-z_1^{2}=z_2^{2} $
Þ $ z_1^{2}+z_2^{2}=z_1z_2 $ Note: Students should remember this question as a formula.
BETA
